Problem: $h(t) = -2t-1$ $f(t) = 6t+4(h(t))$ $ h(f(2)) = {?} $
Answer: First, let's solve for the value of the inner function, $f(2)$ . Then we'll know what to plug into the outer function. $f(2) = (6)(2)+4(h(2))$ To solve for the value of $f$ , we need to solve for the value of $h(2)$ $h(2) = (-2)(2)-1$ $h(2) = -5$ That means $f(2) = (6)(2)+(4)(-5)$ $f(2) = -8$ Now we know that $f(2) = -8$ . Let's solve for $h(f(2))$ , which is $h(-8)$ $h(-8) = (-2)(-8)-1$ $h(-8) = 15$